Problem: Let $f(x)=10^{(x^2-1)}$. Find $f'(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\ln(10)\cdot 10^{(x^2-1)}$ (Choice B) B $(2x)\ln(10)\cdot 10^{(x^2-1)}$ (Choice C) C $(x^2-1)10^{(x^2-2)}$ (Choice D) D $(2x)10^{(x^2-1)}$
$f$ is an exponential function, but its argument isn't simply $x$. Therefore, it's a composite function. In other words, suppose $u(x)=x^2-1$, then $f(x)=10^{u(x)}$. $f'(x)$ can be found using the following identity: $\dfrac{d}{dx}\left[10^{u(x)}\right]=\ln(10)\cdot 10^{u(x)}\cdot u'(x)$ [Why is this identity true?] Let's differentiate! $\begin{aligned} &\phantom{=}f'(x) \\\\ &=\dfrac{d}{dx}10^{(x^2-1)} \\\\ &=\dfrac{d}{dx}10^{u(x)}&&\gray{\text{Let }u(x)=x^2-1} \\\\ &=\ln(10)\cdot 10^{u(x)}\cdot u'(x) \\\\ &=\ln(10)\cdot 10^{(x^2-1)}\cdot (2x)&&\gray{\text{Substitute }u(x)\text{ back}} \\\\ &=(2x)\ln(10)\cdot 10^{(x^2-1)} \end{aligned}$ In conclusion, $f'(x)=(2x)\ln(10)\cdot 10^{(x^2-1)}$.